Zadacha Kuznecov Predely 19-5

Условие задачи

Вычислить предел функции:

$\lim _{x\to 2}\left({\frac {\sin {(3\pi x)}}{\sin {(\pi x)}}}\right)^{\sin ^{2}{(x-2)}}$

Решение

$\lim _{x\to 2}\left({\frac {\sin {(3\pi x)}}{\sin {(\pi x)}}}\right)^{\sin ^{2}{(x-2)}}=\lim _{x\to 2}\left({\frac {\sin {(2\pi x+\pi x)}}{\sin {(\pi x)}}}\right)^{\sin ^{2}{(x-2)}}=$

$=\lim _{x\to 2}\left({\frac {\sin {2\pi x}\cdot \cos {\pi x}+\cos {2\pi x}\cdot \sin {\pi x}}{\sin {(\pi x)}}}\right)^{\sin ^{2}{(x-2)}}=$

$=\lim _{x\to 2}\left({\frac {2\sin {\pi x}\cdot \cos {\pi x}\cdot \cos {\pi x}+\cos {2\pi x}\cdot \sin {\pi x}}{\sin {(\pi x)}}}\right)^{\sin ^{2}{(x-2)}}=$

$=\lim _{x\to 2}\left({\frac {\sin {\pi x}\left(2\cos ^{2}{\pi x}+\cos {2\pi x}\right)}{\sin {(\pi x)}}}\right)^{\sin ^{2}{(x-2)}}=$

$=\lim _{x\to 2}\left(2\cos ^{2}{\pi x}+\cos {2\pi x}\right)^{\sin ^{2}{(x-2)}}=$

$=\left(2\cos ^{2}{(\pi \cdot 2)}+\cos {(2\pi \cdot 2)}\right)^{\sin ^{2}{(2-2)}}=$

$=\left(2\cdot 1^{2}+1\right)^{0^{2}}=3^{0}=1$